Problem: $f(x)=\dfrac{1}{x}$ Find the third degree Taylor polynomial, centered at $x=1$, of $f$. Choose 1 answer: Choose 1 answer: (Choice A) A $1-x+\frac{2{{x}^{2}}}{2!}-\frac{6{{x}^{3}}}{3!}$ (Choice B) B $(x-1)+\frac{2{{(x-1)}^{2}}}{2!}-\frac{6{{(x-1)}^{3}}}{3!}$ (Choice C) C $1-(x-1)+\frac{2{{(x-1)}^{2}}}{2!}-\frac{6{{(x-1)}^{3}}}{3!}$ (Choice D) D $x+\frac{2{{x}^{2}}}{2!}-\frac{6{{x}^{3}}}{3!}$
Explanation: First, find the first three derivatives of $~f(x)\,$. ${f}\,^\prime(x)=-\frac{1}{x^2}~;~~~{f}\,^{\prime\prime}(x)=\frac{2}{x^3}~;~~~{f}\,^{\prime\prime\prime}(x)=-\frac{6}{x^4}$ Then let $~x=1~$ in the original function and in these derivatives to get the coefficients for $~{{T}_{3}}\left( x \right)\,$, the third degree Taylor polynomial for $~f\left( x \right)\,$. $ f(1)=1~;~~~{f}\,^\prime(1)=-1~;~~~{f}\,^{\prime\prime}(1)=2~;~~~{f}\,^{\prime\prime\prime}(1)=-6$ Use these coefficients in the equation for the definition of the Taylor polynomial of degree $~3\,$. ${{T}_{3}}\left( x \right)=f(1)+{f}\,^\prime(1)(x-1)+\frac{f\,^{\prime\prime}(1){{(x-1)}^{2}}}{2!}+\frac{f\,^{\prime\prime\prime}(1){{(x-1)}^{3}}}{3!}$ Hence, ${{T}_{3}}\left( x \right)=1-(x-1)+\frac{2{{(x-1)}^{2}}}{2!}-\frac{6{{(x-1)}^{3}}}{3!}\,$.